Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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− | We have <math>(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!})+\frac{1}{2022!}=(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!})+\frac{1}{2021!}</math> from canceling a 2022 from <math>\frac{2021+1}{2022!}</math>. | + | We have <math>\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}</math> from canceling a 2022 from <math>\frac{2021+1}{2022!}</math>. |
− | This sum clearly telescopes, thus we end up with <math>(\frac{1}{2!}+\frac{2}{3!})+\frac{1}{3!}=\frac{2}{2!}=1</math>. Thus the original equation is equal to <math>1-\frac{1}{2022}</math>, and <math>1+2022=2023</math>. <math>\boxed{\textbf{(D)}\ 2023}</math>. | + | This sum clearly telescopes, thus we end up with <math>\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1</math>. Thus the original equation is equal to <math>1-\frac{1}{2022}</math>, and <math>1+2022=2023</math>. <math>\boxed{\textbf{(D)}\ 2023}</math>. |
~not_slay | ~not_slay |
Revision as of 13:37, 19 November 2022
Problem
The sum
can be expressed as
, where
and
are positive integers. What is
?
Solution 1
Note that , and therefore this sum is a telescoping sum, which is equivalent to
. Our answer is
.
~mathboy100
Solution 2
We have from canceling a 2022 from
.
This sum clearly telescopes, thus we end up with
. Thus the original equation is equal to
, and
.
.
~not_slay
Solution 3 (Induction)
By looking for a pattern, we see that and
, so we can conclude by engineer's induction that the sum in the problem is equal to
, for an answer of
. This can be proven with actual induction as well; we have already established
base cases, so now assume that
for
. For
we get
, completing the proof.
~eibc
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.